# Integrate the following function:

$\sin x\sin \left( {\cos x} \right).$

Last updated date: 21st Mar 2023

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Answer

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Hint: - Substitute the value of $\cos x = t$ and differentiate the equation with respect to x.

Let, $I = \int {\sin x\sin \left( {\cos x} \right)dx} $

Substitute, $\cos x = t.............\left( 1 \right)$

Differentiate equation 1 w.r.t. $x$

As we know differentiation of $\cos x = - \sin x$

$ \Rightarrow - \sin xdx = dt$

Substitute this value in the integral we have

$

I = \int {\sin \left( t \right)\left( { - dt} \right)} \\

\Rightarrow I = - \int {\sin tdt} \\

$

Now as we know integration of $\sin t$ is $- \cos t$

$ \Rightarrow I = - \left( { - \cos t} \right) + c$, where c is some arbitrary integration constant

Now put the value of $t$

$

\Rightarrow I = \cos t + c \\

\Rightarrow I = \cos \left( {\cos x} \right) + c \\

$

So, this is the required value of the integral.

Note: - In such types of question the key concept we have to remember is that always substitute some values to $t$ or any other variable, to make integration simple, then differentiate the variable you assumed w.r.t the given variable, then re-substitute this value in to integral, then always remember the basic differentiation and integration formulas, then simplify we will get the required value of the integral.

Let, $I = \int {\sin x\sin \left( {\cos x} \right)dx} $

Substitute, $\cos x = t.............\left( 1 \right)$

Differentiate equation 1 w.r.t. $x$

As we know differentiation of $\cos x = - \sin x$

$ \Rightarrow - \sin xdx = dt$

Substitute this value in the integral we have

$

I = \int {\sin \left( t \right)\left( { - dt} \right)} \\

\Rightarrow I = - \int {\sin tdt} \\

$

Now as we know integration of $\sin t$ is $- \cos t$

$ \Rightarrow I = - \left( { - \cos t} \right) + c$, where c is some arbitrary integration constant

Now put the value of $t$

$

\Rightarrow I = \cos t + c \\

\Rightarrow I = \cos \left( {\cos x} \right) + c \\

$

So, this is the required value of the integral.

Note: - In such types of question the key concept we have to remember is that always substitute some values to $t$ or any other variable, to make integration simple, then differentiate the variable you assumed w.r.t the given variable, then re-substitute this value in to integral, then always remember the basic differentiation and integration formulas, then simplify we will get the required value of the integral.

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